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人気のある三角関数 >

-1=cos(t)+cos(2t)

解

- 1 = cos (t) + cos (2 t)

解

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

+1

度

t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n, t = 12 0^{\circ} + 36 0^{\circ} n, t = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

- 1 = cos (t) + cos (2 t)

辺を交換する

cos (t) + cos (2 t) = - 1

両辺から

- 1

を引く

cos (t) + cos (2 t) + 1 = 0

三角関数の公式を使用して書き換える

1 + cos (2 t) + cos (t)

2倍角の公式を使用:

cos (2 x) = 2 cos^{2} (x) - 1

= 1 + 2 cos^{2} (t) - 1 + cos (t)

簡素化

1 + 2 cos^{2} (t) - 1 + cos (t) : 2 cos^{2} (t) + cos (t)

1 + 2 cos^{2} (t) - 1 + cos (t)

条件のようなグループ

= 2 cos^{2} (t) + cos (t) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (t) + cos (t)

= 2 cos^{2} (t) + cos (t)

cos (t) + 2 cos^{2} (t) = 0

置換で解く

cos (t) + 2 cos^{2} (t) = 0

仮定:

cos (t) = u

u + 2 u^{2} = 0

u + 2 u^{2} = 0 : u = 0, u = - \frac{1}{2}

u + 2 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

2 u^{2} + u = 0

解くとthe二次式

2 u^{2} + u = 0

二次Equationの公式:

次の場合:

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 1 - 0

数を引く:

1 - 0 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 1 + 1}{2 \cdot 2}, u_{2} = \frac{- 1 - 1}{2 \cdot 2}

u = \frac{- 1 + 1}{2 \cdot 2} : 0

\frac{- 1 + 1}{2 \cdot 2}

数を足す/引く:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{4}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 1 - 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 1 - 1}{2 \cdot 2}

数を引く:

- 1 - 1 = - 2

= \frac{- 2}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{4}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

二次equationの解:

u = 0, u = - \frac{1}{2}

代用を戻す

u = cos (t)

cos (t) = 0, cos (t) = - \frac{1}{2}

cos (t) = 0, cos (t) = - \frac{1}{2}

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

以下の一般解

cos (t) = 0

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = - \frac{1}{2}

以下の一般解

cos (t) = - \frac{1}{2}

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

すべての解を組み合わせる

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

グラフ

人気の例

15 cos (3 x) = 0

- \frac{1}{2} = sin (2 x)

72(-3cos^2(x)sin^2(x)+cos^4(x))=0

72 (- 3 cos^{2} (x) sin^{2} (x) + cos^{4} (x)) = 0

sin(x/5)=(sqrt(2))/2

sin (\frac{x}{5}) = \frac{2}{2}

solvefor x,sin(x)=0.6

so l v e f or x, sin (x) = 0.6